# Solution Manual For General Chemistry: Principles and Modern Applications, 11th Edition

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CHAPTER 2
ATOMS AND THE ATOMIC THEORY
PRACTICE EXAMPLES
1A
(E) The total mass must be the same before and after reaction.
mass before reaction = 0.382 g magnesium + 2.652 g nitrogen = 3.034 g
mass after reaction = magnesium nitride mass + 2.505 g nitrogen = 3.034 g
magnesium nitride mass = 3.034 g ๏ญ 2.505 g = 0.529 g magnesium nitride
1B
(E) Again, the total mass is the same before and after the reaction.
mass before reaction = 7.12 g magnesium +1.80 g bromine = 8.92 g
mass after reaction = 2.07 g magnesium bromide + magnesium mass = 8.92 g
magnesium mass = 8.92 g ๏ญ 2.07 g = 6.85 g magnesium
2A
(M) In Example 2-2 we are told that 0.500 g MgO contains 0.301 g of Mg. With this
information, we can determine the mass of magnesium needed to form 2.000 g of
magnesium oxide.
mass of Mg = 2.000 g MgO ๏ด
0.301 g Mg
= 1.20 g Mg
0.500 g MgO
The remainder of the 2.00 g of magnesium oxide is the mass of oxygen.
mass of oxygen = 2.00 g magnesium oxide ๏ญ 1.20 g magnesium = 0.80 g oxygen
2B
(M) In Example 2-2, we see that a 0.500 g sample of MgO has 0.301 g Mg, hence, it must
have 0.199 g O2. From this we see that if we have equal masses of Mg and O2, the oxygen is
in excess. First we find out how many grams of oxygen reacts with 10.00 g of Mg.
0.199 g O 2
๏ฝ 6.61 g O 2 (used up)
0.301 g Mg
Hence, 10.00 g ๏ญ 6.61 g ๏ฝ 3.39 g O 2 unreacted. Mg is the limiting reactant.
mass oxygen ๏ฝ 10.00 g Mg ๏ด
MgO(s) mass ๏ฝ mass Mg + Mass O 2 ๏ฝ 10.00 g ๏ซ 6.61 g ๏ฝ 16.61 g MgO.
There are only two substances present, 16.61 g of MgO (product) and 3.39 g of unreacted O2
3A
(E) Silver has 47 protons. If the isotope in question has 62 neutrons, then it has a mass
number of 109. This can be represented as 109
47 Ag .
3B
(E) Tin has 50 electrons and 50 protons when neutral, while a neutral cadmium atom has
48 electrons. This means that we are dealing with Sn2+. We do not know how many
neutrons tin has. so there can be more than one answer. For instance,
116
2+ 117
2+ 118
2+ 119
2+
120
2+
are all possible answers.
50 Sn , 50 Sn , 50 Sn , 50 Sn , and 50 Sn
4A
(E) The ratio of the masses of 202Hg and 12C is:
202
12
Hg 201.97062 u
๏ฝ
๏ฝ 16.830885
C
12 u
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Chapter 2: Atoms and the Atomic Theory
4B
(E) Atomic mass is 12 u ร 13.16034 = 157.9241 u. The isotope is 158
64 Gd . Using an atomic
16
mass of 15.9949 u for 16O, the mass of 158
O is
64 Gd relative to
relative mass to oxygen-16 =
157.9241 u
๏ฝ 9.87340
15.9949 u
5A
(E) The atomic mass of boron is 10.81, which is closer to 11.0093054 than to 10.0129370.
Thus, boron-11 is the isotope that is present in greater abundance.
5B
(E) The average atomic mass of indium is 114.82, and one isotope is known to be 113In.
Since the weighted-average atomic mass is almost 115, the second isotope must be larger
than both In-113 and In-114. Clearly, then, the second isotope must be In-115 (115In). Since
the average atomic mass of indium is closest to the mass of the second isotope, In-115, then
115
In is the more abundant isotope.
6A
(M) Weighted-average atomic mass of Si =
(27.9769265325 u ร 0.9223) ๏ฎ 25.80 u
(28.976494700 u ร 0.04685) ๏ฎ 1.358 u
(29.973377017 u ร 0.03092) ๏ฎ 0.9268 u
28.0848 u
We should report the weighted-average atomic mass of Si as 28.08 u.
6B
(M) We let x be the fractional abundance of lithium-6.
6.941 u = ๏ x ๏ด 6.01512 u ๏ + ๏ฉ๏ซ๏จ1 ๏ญ x ๏ฉ ๏ด 7.01600 u ๏น๏ป = x ๏ด 6.01512 u + 7.01600 u ๏ญ x ๏ด 7.01600 u
6.941 u ๏ญ 7.01600 u = x ๏ด 6.01512 u ๏ญ x ๏ด 7.01600 u = ๏ญ x ๏ด1.00088 u
x๏ฝ
7A
6.941 u ๏ญ 7.01600 u
๏ฝ 0.075 Percent isotopic abundances : 7.5% lithium-6, 92.5% lithium-7
๏ญ1.00088 u
(M) We assume that atoms lose or gain relatively few electrons to become ions. Thus,
elements that will form cations will be on the left-hand side of the periodic table, while
elements that will form anions will be on the right-hand side. The number of electrons
โlostโ when a cation forms is usually equal to the last digit of the periodic group number;
the number of electrons added when an anion forms is typically eight minus the last digit
of the group number.
Li is in group 1(1A); it should form a cation by losing one electron: Li + .
S is in group 6(6A); it should form an anion by adding two electrons: S2๏ญ .
Ra is in group 2(2A); it should form a cation by losing two electrons: Ra 2+ .
F and I are both group 17(7A); they should form anions by gaining an electron: F ๏ญ and I ๏ญ .
A1 is in group 13(3A); it should form a cation by losing three electrons: Al 3+ .
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Chapter 2: Atoms and the Atomic Theory
7B
8A
(M) Main-group elements are in the โAโ families, while transition elements are in the โBโ
families. Metals, nonmetals, metalloids, and noble gases are color coded in the periodic
table inside the front cover of the textbook.
Na is a main-group metal in
group 1(1A).
Re is a transition metal in group 7(7B).
S is a main-group nonmetal in
group 16(6A).
I is a main-group nonmetal in group 17(7A).
Kr is a nonmetal in group 18(8A).
Mg is a main-group metal in group 2(2A).
U is an inner transition metal,
an actinide.
Si is a main-group metalloid in group 14(4A).
B is a metalloid in group 13(3A).
A1 is a main-group metal in group 13(3A).
As is a main-group metalloid in
group 15(5A).
H is a main-group nonmetal in group 1(1A).
(H is believed to be a metal at extremely high
pressures.)
(E) This is similar to Practice Examples 2-8A and 2-8B.
Cu mass = 2.35 ๏ด1024 Cu atoms ๏ด
8B
(M) Of all lead atoms, 24.1% are lead-206, or 241 206 Pb atoms in every 1000 lead atoms.
First we need to convert a 22.6 gram sample of lead into moles of lead (below) and then,
by using Avogadroโs constant, and the percent isotopic abundance, we can determine the
number of 206 Pb atoms.
1 mol Pb
nPb = 22.6 g Pbร
๏ฝ 0.109 mol Pb
207.2 g Pb
206
9A
Pb atoms = 0.109 mol Pbร
206
6.022ร1023 Pb atoms 241 Pb atoms
ร
= 1.58ร10 22 206 Pb atoms
1mol Pb
1000 Pb atoms
(M) Both the density and the molar mass of Pb serve as conversion factors.
atoms of Pb = 0.105cm3 Pb ๏ด
9B
1 mol Cu
63.546 g Cu
๏ด
= 248 g Cu
23
6.022 ๏ด10 atoms
1 mol Cu
11.34 g 1 mol Pb
1 cm
๏ด
3
207.2 g
๏ด
6.022 ๏ด 10 23 Pb atoms
1mol Pb
= 3.46 ๏ด 1021 Pb atoms
(M) First we find the number of rhenium atoms in 0.100 mg of the element.
1g
1 mol Re
6.022 ๏ด 1023 Re atoms
0.100 mg ๏ด
๏ด
๏ด
๏ฝ 3.23 ๏ด 1017 Re atoms
1000 mg 186.21 g Re
1 mol Re
% abundance
187
2.02 ๏ด 1017 atoms 187 Re
Re =
๏ด 100% = 62.5%
3.23 ๏ด 1017 Re atoms
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Chapter 2: Atoms and the Atomic Theory
INTEGRATIVE EXAMPLE
A.
(M)
Stepwise approach:
First, determine the total number of Cu atoms in the crystal by determining the volume of
the crystal and calculating the mass of Cu from density. Then we can determine the amount
of 63Cu by noting its relative abundance
1 cm3
3
๏ฝ 1.5625 ๏ด 10 ๏ญ17 cm3
Volume of crystal = ๏จ 25 nm ๏ฉ ๏ด
7
3
(1 ๏ด 10 nm)
Mass of Cu in crystal = d ๏ V = 8.92 g/cm3 ร 1.5625 ร 10โ17 = 1.3938 ร 10โ16 g
# of Cu atoms =
1 mol Cu
6.022 ๏ด 1023 Cu atoms
1.3938 ๏ด 10 g Cu ๏ด
๏ด
๏ฝ 1.3208 ๏ด 106 Cu atoms
63.546 g Cu
1 mol Cu
63
Therefore, the number of Cu atoms, assuming 69.17% abundance, is 9.14 ร 105 atoms.
๏ญ16
Conversion pathway approach:
8.92 g Cu
1 cm3
(25 nm)3
1 mol Cu
6.022 ๏ด 1023 Cu atoms
๏ด
๏ด
๏ด
๏ด
1 cm3
(1 ๏ด 107 nm)3
crystal
63.546 g Cu
1 mol Cu
69.17 atoms of 63Cu
๏ด
๏ฝ 9.14 ๏ด 105 atoms of 63Cu
100 Cu atoms
B.
(M)
Stepwise approach:
Calculate the mass of Fe in a serving of cereal, determine mass of 58Fe in that amount of
cereal, and determine how many servings of cereal are needed to reach 58 g of 58Fe.
Amount of Fe in a serving of cereal = 18 mg ร 0.45 = 8.1 mg Fe per serving
First calculate the amount of Fe
1 mol Fe
0.0081 g Fe ๏ด
๏ฝ 1.45 ๏ด10๏ญ4 mol Fe
55.845 g Fe
Then calculate 58Fe amount:
0.282 mol 58 Fe
โ4
= 4.090ร10โ7 mol 58Fe
1.45 ร 10 mol Fe ร
100 mol Fe
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Chapter 2: Atoms and the Atomic Theory
Converting mol of 58F to # of servings:
4.090 ๏ด 10 ๏ญ7 mol 58 Fe 57.9333 g 58 Fe
๏ด
๏ฝ 2.37 ๏ด 10 ๏ญ5 g 58 Fe per serving
58
1 serving
1 mol Fe
Total # of servings = 58 g total / 2.37ร10โ5 per serving = 2.4477ร106 servings
Conversion Pathway Approach:
The number of servings of dry cereal to ingest 58 g of 58Fe =
1 mol 58 Fe
100 mol Fe
58.845 g Fe 1 cereal serving
58.0 g 58 Fe ๏ด
๏ด
๏ด
๏ด
58
58
57.9333 g Fe 0.282 mol Fe
1 mol Fe
0.018 g Fe ๏ด 0.45
๏ฝ 2.4477 ๏ด 106 servings
2.44477 ๏ด 106 servings ๏ด
1 year
๏ฝ 6706 years
365 servings
EXERCISES
Law of Conservation of Mass
1.
(E) The observations cited do not necessarily violate the law of conservation of mass. The
oxide formed when iron rusts is a solid and remains with the solid iron, increasing the mass
of the solid by an amount equal to the mass of the oxygen that has combined. The oxide
formed when a match burns is a gas and will not remain with the solid product (the ash); the
mass of the ash thus is less than that of the match. We would have to collect all reactants
and all products and weigh them to determine if the law of conservation of mass is obeyed
or violated.
2.
(E) The magnesium that is burned in air combines with some of the oxygen in the air and this
oxygen (which, of course, was not weighed when the magnesium metal was weighed) adds its
mass to the mass of the magnesium, making the magnesium oxide product weigh more than
did the original magnesium. When this same reaction is carried out in a flashbulb, the oxygen
(in fact, some excess oxygen) that will combine with the magnesium is already present in the
bulb before the reaction. Consequently, the product contains no unweighed oxygen.
3.
(E) By the law of conservation of mass, all of the magnesium initially present and all of the
oxygen that reacted are present in the product. Thus, the mass of oxygen that has reacted is
obtained by difference. mass of oxygen = 0.674 g MgO ๏ญ 0.406 g Mg = 0.268 g oxygen
4.
(E) Reaction: 2 K(s) + Cl2(g) ๏ฎ 2 KCl(s)
Mass of Cl2 reacted = 8.178 g โ 6.867 g = 1.311 g Cl2(g)
mKCl ๏ฝ 1.311 g Cl2 ๏ด
1 mol Cl2 2 mol KCl 74.55 g KCl
๏ด
๏ด
๏ฝ 2.757 g KCl
70.90 g Cl2 1 mol Cl2
1 mol KCl
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Chapter 2: Atoms and the Atomic Theory
5.
(M) We need to compare the mass before reaction (initial) with that after reaction (final) to
answer this question.
initial mass = 10.500 g calcium hydroxide +11.125 g ammonium chloride = 21.625 g
final mass = 14.336 g solid residue + (69.605 โ 62.316) g of gases = 21.625 g
These data support the law of conservation of mass. Note that the gain in the mass of water
is equal to the mass of gas absorbed by the water.
6.
(M) We compute the mass of the reactants and compare that with the mass of the products
to answer this question.
reactant mass = mass of calcium carbonate + mass of hydrochloric acid solution
1.148g
= 10.00 g calcium carbonate +100.0 mL soln ๏ด
1 mL soln
= 10.00 g calcium carbonate +114.8g solution = 124.8 g reactants
product mass = mass of solution + mass of carbon dioxide
1.9769 g
= 120.40 g soln + 2.22 L gas ๏ด
1 L gas
= 120.40 g soln + 4.39 g carbon dioxide
๏ฆ The same mass within experimental error, ๏ถ
= 124.79 g products ๏ง
๏ท
๏จ thus, the law of conservation of mass obeyed. ๏ธ
Law of Constant Composition
7.
(E)
(a)
(b)
(c)
8.
(0.755 ๏ญ 0.455) g
๏ฝ 0.397
0.755 g
0.300 g
Ratio of O: Mg in MgO by mass ๏ฝ
๏ฝ 0.659
0.455 g
0.455 g Mg
Percent magnesium by mass ๏ฝ
ร100% ๏ฝ 60.3%
0.755 g MgO
Ratio of O: MgO by mass ๏ฝ
(M)
(a) We can determine that carbon dioxide has a fixed composition by finding the % C
in each sample. (In the calculations below, the abbreviation โcmpdโ is short for
compound.)
3.62 g C
5.91 g C
%C ๏ฝ
๏ด 100% ๏ฝ 27.3% C
%C ๏ฝ
๏ด 100% ๏ฝ 27.3% C
13.26 g cmpd
21.66 g cmpd
7.07 g C
%C ๏ฝ
๏ด 100% ๏ฝ 27.3% C
25.91 g cmpd
Since all three samples have the same percent of carbon, these data do establish that
carbon dioxide has a fixed composition.
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Chapter 2: Atoms and the Atomic Theory
(b)
9.
Carbon dioxide contains only carbon and oxygen. As determined in (a), carbon dioxide
is 27.3% C by mass. The percent of oxygen in carbon dioxide is obtained by difference.
%O = 100.0 % โ (27.3 %C) = 72.7 %O
(M) In the first experiment, 2.18 g of sodium produces 5.54 g of sodium chloride. In the
second experiment, 2.10 g of chlorine produces 3.46 g of sodium chloride. The amount of
sodium contained in this second sample of sodium chloride is given by
mass of sodium = 3.46 g sodium chloride ๏ญ 2.10 g chlorine = 1.36 g sodium.
We now have sufficient information to determine the % Na in each of the samples of
sodium chloride.
2.18g Na
1.36 g Na
%Na =
๏ด100% = 39.4% Na
%Na =
๏ด100% = 39.3% Na
5.54 g cmpd
3.46 g cmpd
Thus, the two samples of sodium chloride have the same composition. Recognize that,
based on significant figures, each percent has an uncertainty of ๏ฑ0.1% .
10.
(E) If the two samples of water have the same % H, the law of constant composition is
demonstrated. Notice that, in the second experiment, the mass of the compound is equal to
the sum of the masses of the elements produced from it.
3.06 g H
1.45 g H
%H =
๏ด 100% ๏ฝ 11.2% H
%H ๏ฝ
๏ด 100% ๏ฝ 11.2% H
27.35g H 2 O
๏จ1.45 +11.51๏ฉ g H 2O
Thus, the results are consistent with the law of constant composition.
11.
(E) The mass of sulfur (0.312 g) needed to produce 0.623 g sulfur dioxide provides the
information required for the conversion factor.
0.312 g sulfur
sulfur mass = 0.842 g sulfur dioxide ๏ด
= 0.422 g sulfur
0.623g sulfur dioxide
12.
(M)
(a) From the first experiment we see that 1.16 g of compound is produced per gram
of Hg. These masses enable us to determine the mass of compound produced from
1.50 g Hg.
1.16 g cmpd
mass of cmpd = 1.50 g Hg ๏ด
= 1.74 g cmpd
1.00 g Hg
(b)
Since the compound weighs 0.24 g more than the mass of mercury (1.50 g) that was
used, 0.24 g of sulfur must have reacted. Thus, the unreacted sulfur has a mass of
0.76 g (= 1.00 g initially present ๏ญ๏ ๏ฐ๏ฎ24 g reacted).
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Chapter 2: Atoms and the Atomic Theory
Law of Multiple Proportions
13.
(M) By dividing the mass of the oxygen per gram of sulfur
in the second sulfur-oxygen compound (compound 2) by the
mass of oxygen per gram of sulfur in the first sulfur-oxygen
compound (compound 1), we obtain the ratio (shown to the
right):
1.497 g of O
(cmpd 2)
1.500
1.000 g of S
๏ฝ
0.998 g of O
1
(cmpd 1)
1.000 g of S
To get the simplest whole number ratio we need to multiply both the numerator and the
denominator by 2. This gives the simple whole number ratio 3/2. In other words, for a given
mass of sulfur, the mass of oxygen in the second compound (SO3) relative to the mass of
oxygen in the first compound (SO2) is in a ratio of 3:2. These results are entirely consistent
with the Law of Multiple Proportions because the same two elements, sulfur and oxygen in
this case, have reacted together to give two different compounds that have masses of oxygen
that are in the ratio of small positive integers for a fixed amount of sulfur.
14.
(M) This question is similar to question 13 in that two elements, phosphorus and chlorine in
this case, have combined to give two different compounds. This time, however, different
masses have been used for both of the elements in the second compound. To see if the Law
of Multiple Proportions is being followed, the mass of one of the two elements must be set
to the same value in both compounds. This can be achieved by dividing the masses of both
phosphorus and chlorine in reaction 2 by 2.500:
2.500 g phosphorus
โnormalizedโ mass of phosphorus =
= 1.000 g of phosphorus
2.500
14.308 g chlorine
= 5.723 g of chlorine
โnormalizedโ mass of chlorine =
2.500
Now the mass of phosphorus for both reactions is fixed at 1.000 g. Next, we will divide each
amount of chlorine by the fixed mass of phosphorus with which they are combined. This gives
3.433 g of Cl
(cmpd 1)
1.000 g P
๏ฝ 0.600 ๏ฝ 6 :10 or 3 : 5
5.723 g of Cl
(cmpd 2)
1.000 g P
15.
(M)
(a) First of all we need to fix the mass of nitrogen in all three compounds to some
common value, for example, 1.000 g. This can be accomplished by multiplying the
masses of hydrogen and nitrogen in compound A by 2 and the amount of hydrogen
and nitrogen in compound C by 4/3 (1.333):
Cmpd A โnormalizedโ mass of nitrogen
โnormalizedโ mass of hydrogen
= 0.500 g N ๏ด 2 = 1.000 g N
= 0.108 g H ๏ด 2 = 0.216 g H
Cmpd C โnormalizedโ mass of nitrogen
โnormalizedโ mass of hydrogen
= 0.750 g N ๏ด 1.333 = 1.000 g N
= 0.108 g H ๏ด 1.333 = 0.144 g H
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Chapter 2: Atoms and the Atomic Theory
Next, we divide the mass of hydrogen in each compound by the smallest mass of
hydrogen, namely, 0.0720 g. This gives 3.000 for compound A, 1.000 for compound
B, and 2.000 for compound C. The ratio of the amounts of hydrogen in the three
compounds is 3 (cmpd A) : 1 (cmpd B) : 2 (cmpd C)
These results are consistent with the Law of Multiple Proportions because the masses
of hydrogen in the three compounds end up in a ratio of small whole numbers when the
mass of nitrogen in all three compounds is normalized to a simple value (1.000 g here).
(b)
16.
The text states that compound B is N2H2. This means that, based on the relative
amounts of hydrogen calculated in part (a), compound A might be N2H6 and
compound C, N2H4. Actually, compound A is NH3, but we have no way of knowing
this from the data. Note that the H:N ratios in NH3 and N2H6 are the same, 3H:1N.
(M)
(a) As with the previous problem, one of the two elements must have the same mass in all
of the compounds. This can be most readily achieved by setting the mass of iodine in
all four compounds to 1.000 g. With this approach we only need to manipulate the data
for compounds B and C. To normalize the amount of iodine in compound B to 1.000 g,
we need to multiply the masses of both iodine and fluorine by 2. To accomplish the
analogous normalization of compound C, we must multiply by 4/3 (1.333).
Cmpd B: โnormalizedโ mass of iodine = 0. 500 g I ๏ด 2
= 1.000 g I
โnormalizedโ mass of fluorine = 0.2246 g F ๏ด 2
= 0.4492 g F
Cmpd C: โnormalizedโ mass of iodine = 0.750 g I ๏ด 1.333 = 1.000 g I
โnormalizedโ mass of fluorine = 0.5614 g F ๏ด 1.333 = 0.7485 g F
Next we divide the mass of fluorine in each compound by the smallest mass of fluorine,
namely, 0.1497 g. This gives 1.000 for compound A, 3.001 for compound B, 5.000 for
compound C, and 7.001 for compound D. The ratios of the amounts of fluorine in the
four compounds A : B : C : D is 1 : 3 : 5 : 7. These results are consistent with the law
of multiple proportions because for a fixed amount of iodine (1.000 g), the masses of
fluorine in the four compounds are in the ratio of small whole numbers.
(b)
17.
As with the preceding problem, we can figure out the empirical formulas for the four
iodine-fluorine containing compounds from the ratios of the amounts of fluorine that
were determined in 16(a): Cmpd A: IF Cmpd B: IF3 Cmpd C: IF5 Cmpd D: IF7
(M) One oxide of copper has about 20% oxygen by mass. If we assume a 100 gram sample,
then ~ 20 grams of the sample is oxygen (~1.25 moles) and 80 grams is copper (~1.26 moles).
This would give an empirical formula of CuO (copper(II) oxide). The second oxide has less
oxygen by mass, hence the empirical formula must have less oxygen or more copper (Cu:O
ratio greater than 1). If we keep whole number ratios of atoms, a plausible formula would be
Cu2O (copper(I) oxide), where the mass percent oxygen is ๏ป11%.
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Chapter 2: Atoms and the Atomic Theory
18.
(M) Assuming the intermediate is โhalf-wayโ between CO (oxygen-carbon mass ratio = 16:12
or 1.333) and CO2 (oxygen-carbon mass ratio = 32:12 or 2.6667), then the oxygen- carbon
ratio would be 2:1, or O:C = 24:12. This mass ratio gives a mole ratio of O:C = 1.5:1.
Empirical formulas are simple whole number ratios of elements; hence, a formula of C3O2
must be the correct empirical formula for this carbon oxide. (Note: C3O2 is called tricarbon
dioxide or carbon suboxide).
Fundamental Charges and Mass-to-Charge Ratios
19.
(M) We can calculate the charge on each drop, express each in terms of 10 ๏ญ19 C, and finally
express each in terms of e = 1.6 ๏ด10๏ญ19 C.
๏ฝ 12.8 ๏ด10๏ญ19 C
= 8e
drops 2 &3 : 1.28 ๏ด10๏ญ18 ๏ธ 2 ๏ฝ 0.640 ๏ด10๏ญ18 C ๏ฝ 6.40 ๏ด10๏ญ19 C
= 4e
drop 4 :
1.28 ๏ด10๏ญ18 ๏ธ 8 ๏ฝ 0.160 ๏ด10๏ญ18 C ๏ฝ 1.60 ๏ด10๏ญ19 C
= 1e
drop 5 :
1.28 ๏ด10๏ญ18 ๏ด 4 ๏ฝ 5.12 ๏ด 10๏ญ18 C
= 32 e
drop 1:
1.28 ๏ด10๏ญ18
๏ฝ 51.2 ๏ด10๏ญ19 C
We see that these values are consistent with the charge that Millikan found for that of the
electron, and he could have inferred the correct charge from these data, since they are all
multiples of e .
20.
(M) We calculate each dropโs charge, express each in terms of 10 ๏ญ19 C, and then, express
each in terms of e = 1.6 ๏ด10๏ญ19 C.
drop 1: 6.41๏ด 10๏ญ19 C
๏ฝ 6.41๏ด 10๏ญ19 C
= 4e
drop 2 : 6.41๏ด 10๏ญ19 ๏ธ 2
๏ฝ 3.21๏ด 10๏ญ19 C
๏ฝ 3.21๏ด10๏ญ19 C = 2 e
drop 3 : 6.41๏ด 10๏ญ19 ๏ด 2
๏ฝ 1.28 ๏ด10๏ญ18 C
๏ฝ 12.8 ๏ด10๏ญ19 C = 8 e
drop 4 : 1.44 ๏ด10๏ญ18
๏ฝ 14.4 ๏ด 10๏ญ19 C
drop 5 : 1.44 ๏ด 10
๏ญ18
๏ธ3
๏ฝ 4.8 ๏ด 10
๏ญ19
= 9e
๏ฝ 4.8 ๏ด10
C
๏ญ19
C
= 3e
We see that these values are consistent with the charge that Millikan found for that of the
electron. He could have inferred the correct charge from these values, since they are all
multiples of e , and have no other common factor.
21.
(M)
(a) Determine the ratio of the mass of a hydrogen atom to that of an electron. We use the
mass of a proton plus that of an electron for the mass of a hydrogen atom.
or
mass of proton ๏ซ mass of electron 1.0073u ๏ซ 0.00055 u
๏ฝ
๏ฝ 1.8 ๏ด 103
mass of electron
0.00055 u
mass of electron
1
๏ฝ
๏ฝ 5.6 ๏ด 10๏ญ4
mass of proton ๏ซ mass of electron 1.8 ๏ด 103
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Chapter 2: Atoms and the Atomic Theory
(b)
The only two mass-to-charge ratios that we can determine from the data in Table 2-1
are those for the proton (a hydrogen ion, H+) and the electron.
mass 1.673 ๏ด10๏ญ24 g
For the proton :
=
= 1.044 ๏ด10๏ญ5 g/C
๏ญ19
charge 1.602 ๏ด 10 C
mass
9.109 ๏ด10๏ญ28 g
=
= 5.686 ๏ด 10๏ญ9 g/C
๏ญ19
charge 1.602 ๏ด 10 C
The hydrogen ion is the lightest positive ion available. We see that the mass-to-charge
ratio for a positive particle is considerably larger than that for an electron.
For the electron :
22.
(M) We do not have the precise isotopic masses for the two ions. The values of the mass-tocharge ratios are only approximate. This is because some of the mass is converted to energy
(binding energy), that holds all of the positively charged protons in the nucleus together.
Consequently, we have used a three-significant-figure mass for a nucleon, rather than the
more precisely known proton and neutron masses. (Recall that the term โnucleonโ refers to a
nuclear particleโ either a proton or a neutron.)
m 127 nucleons
1 electron
1.67 ๏ด10๏ญ24 g
127 ๏ญ
I
=
= 1.32 ๏ด 10๏ญ3 g/C (7.55 ๏ด 102 C/g)
๏ด
๏ด
๏ญ19
1 electron
1.602 ๏ด10 C 1 nucleon
e
m 32 nucleons
1 electron
1.67 ๏ด 10๏ญ24 g
32 2 ๏ญ
S
=
= 1.67 ๏ด10๏ญ4 g/C (6.00 ๏ด 103C/g)
๏ด
๏ด
๏ญ19
e 2 electrons 1.602 ๏ด 10 C 1 nucleon
Atomic Number, Mass Number, and Isotopes
23.
32
59
226
(E) (a) cobalt-60 60
27 Co (b) phosphorus-32 15 P (c) iron-59 26 Fe (d) radium-226 88 Ra
24.
(E) The nucleus of 202
80 Hg contains 80 protons and (202 โ 80) = 122 neutrons.
Thus, the percent of nucleons that are neutrons is given by
122 neutrons
% neutrons =
๏ด 100 = 60.4% neutrons
202 nucleons
25.a Name
Symbol
Na
Si
Rb
Number of
Protons
11
14
37
Number of
Electrons
11
14a
37a
Number of
Neutrons
12
14
48
Mass
Number
23
28
85
(E) Sodium
Silicon
23
11
28
14
85
37
Potassium
40
19
K
19
19
21
40
Arsenic a
75
33
As
33a
33
42
75
Neon
20
10
Ne 2+
10
8
10
20
80
35
Br
35
35
45
80
82
82
126
208
Rubidium
Bromine
Lead b
b
208
82
Pb
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Chapter 2: Atoms and the Atomic Theory
a
This result assumes that a neutral atom is involved.
Insufficient data. Does not characterize a specific nuclide; several possibilities exist.
The minimum information needed is the atomic number (or some way to obtain it, such
as from the name or the symbol of the element involved), the number of electrons (or some
way to obtain it, such as the charge on the species), and the mass number (or the number
of neutrons).
b
26.
27.
(E)
(a)
Since all of these species are neutral atoms, the numbers of electrons are the atomic
numbers, the subscript numbers. The symbols must be arranged in order of increasing
39
58
59
120
112
122
value of these subscripts. 40
18 Ar < 19 K < 27 Co < 29 Cu < 48 Cd < 50 Sn < 52Te
(b)
The number of neutrons is given by the difference between the mass number and the
atomic number, A-Z. This is the difference between superscripted and subscripted
values and is provided (in parentheses) after each element in the following list.
39
40
59
58
112
122
120
19 K(20) < 18 Ar(22) < 29 Cu(30) < 27 Co(31) < 50 Sn(62) < 52Te(70) < 48 Cd(72)
(c)
Here the nuclides are arranged by increasing mass number, given by the superscripts.
39
40
58
59
112
120
122
19 K < 18 Ar < 27 Co < 29 Cu < 50 Sn < 48 Cd 50% more
that is at least 2.5 times greater than the atomic number. Only 90
neutrons than protons (number of neutrons = 226 โ 90 = 136; number of protons = 90;
90 ร 1.5 = 135, less than the total number of neutrons). 124Sn is close, having
74 neutrons, and 50 protons, so 50 ร 1.5 = 75, slightly less than 50%.
39
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Chapter 2: Atoms and the Atomic Theory
35.
(E)
If we let n represent the number of neutrons and p represent the number of protons,
then p + 4 = n. The mass number is the sum of the number of protons and the number
of neutrons: p + n = 44. Substitution of n = p + 4 yields p + p + 4 = 44. From this relation,
we see p = 20. Reference to the periodic table indicates that 20 is the atomic number of the
element calcium.
36.
(M) We will use the same type of strategy and the same notation as we used previously in
Equation 35 to come up with the answer.
n=p+1
There is one more neutron than the number of protons.
n + p = 9 ร 3 = 27
The mass number equals nine times the ionโs charge of 3+.
Substitute the first relationship into the second, and solve for p.
27 ๏ฝ ( p ๏ซ 1) ๏ซ p ๏ฝ 2 p ๏ซ 1
27 ๏ญ 1
๏ฝ 13
p๏ฝ
2
27
Thus this is the 3+ action of the isotope Al-27 ๏ฎ 13
Al3+ .
37.
(M) The number of protons is the same as the atomic number for iodine, which is 53. There
is one more electron than the number of protons because there is a negative charge on the
ion. Therefore the number of electrons is 54. The number of neutrons is equal to 70, mass
number minus atomic number.
38.
(E) For iodine, Z = 53, and so it has 53 protons. Because the overall charge on the ion is 1โ,
there are 54 electrons in a single ion of iodine-131. The number of neutrons is
131 โ 53 = 78.
39.
(E) For americium, Z = 95. There are 95 protons, 95 electrons, and 241 โ 95 = 146 neutrons
in a single atom of americium-241.
40.
(E) For cobalt, Z = 27. There are 27 protons, 27 electrons, and 60 โ 27 = 33 neutrons in a
single atom of cobalt-60.
Atomic Mass Units, Atomic Masses
41.
(E) There are no chlorine atoms that have a mass of 35.45 u. The masses of individual
chlorine atoms are close to integers and this mass is about midway between two integers.
It is an average atomic mass, the result of averaging two (or more) isotopic masses, each
weighted by its isotopic abundance.
40
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Chapter 2: Atoms and the Atomic Theory
42.
(E) It is exceedingly unlikely that another nuclide would have an exact integral mass. The
mass of carbon-12 is defined as precisely 12 u. Each nuclidic mass is close to integral,
but none that we have encountered in this chapter are precisely integral. The reason is that
each nuclide is composed of protons, neutrons, and electrons, none of which have integral
masses, and there is a small quantity of the mass of each nucleon (nuclear particle) lost in
the binding energy holding the nuclides together. It would be highly unlikely that all of
these contributions would add up to a precisely integral mass.
43.
(E) To determine the weighted-average atomic mass, we use the following expression:
average atomic mass = ๏ฅ ๏จ isotopic mass ๏ด fractional isotopic abundance๏ฉ
Each of the three percents given is converted to a fractional abundance by dividing it by 100.
Mg atomic mass = ๏จ 23.985042 u ๏ด 0.7899 ๏ฉ + ๏จ 24.985837 u ๏ด 0.1000 ๏ฉ + ๏จ 25.982593u ๏ด 0.1101๏ฉ
= 18.95 u + 2.499 u + 2.861u = 24.31 u
44.
(E) To determine the average atomic mass, we use the following expression:
average atomic mass = ๏ฅ ๏จ isotopic mass ๏ด fractional isotopic abundance๏ฉ
Each of the three percents given is converted to a fractional abundance by dividing it by 100.
Cr atomic mass = ๏จ 49.9461๏ด 0.0435 ๏ฉ + ๏จ 51.9405 ๏ด 0.8379 ๏ฉ + ๏จ 52.9407 ๏ด 0.0950 ๏ฉ + ๏จ 53.9389 ๏ด 0.0236 ๏ฉ
= 2.17 u + 43.52 u + 5.03u +1.27 u = 51.99 u
If all digits are carried and then the answer is rounded at the end, the answer is 52.00 u.
45.
(E) We will use the expression to determine the weighted-average atomic mass.
107.868 u = ๏จ106.905092 u ๏ด 0.5184 ๏ฉ + ๏จ 109 Ag ๏ด 0.4816 ๏ฉ = 55.42 u + 0.4816 109 Ag
107.868 u ๏ญ 55.42 u ๏ฝ 0.4816 109 Ag = 52.45 u
46.
109
Ag ๏ฝ
52.45 u
๏ฝ 108.9 u
0.4816
(M) The percent abundance of the two isotopes must add to 100.00%, since there are only
two naturally occurring isotopes of gallium. Thus, we can determine the percent abundance
of the second isotope by using difference:
second isotope = 100.00% โ 60.11% = 39.89%
From the periodic table, we see that the atomic mass of gallium is 69.723 u. We use this
value in the expression to determine the weighted-average atomic mass, along with the
isotopic mass of 69Ga and the fractional abundances of the two isotopes (the percent
abundances divided by 100).
69.723 u ๏ฝ (0.6011 ๏ด 68.925581 u) ๏ซ (0.3989 ๏ด other isotope) ๏ฝ 41.43 u ๏ซ (0.3989 ๏ด other isotope)
69.723 ๏ญ 41.43 u
other isotope ๏ฝ
๏ฝ 70.92 u ๏ฝ mass of 71 Ga, the other isotope
0.3989
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Chapter 2: Atoms and the Atomic Theory
47.
(M) Since the three percent abundances total 100%, the percent abundance of 40 K is found
by difference. % 40 K = 100.0000% ๏ญ 93.2581% ๏ญ 6.7302% = 0.0117%
Then the expression for the weighted-average atomic mass is used, with the percent
abundances converted to fractional abundances by dividing by 100. Note that the average
atomic mass of potassium is 39.0983 u.
๏จ
39.0983u = ๏จ 0.932581 ๏ด 38.963707 u ๏ฉ + ๏จ 0.000117 ๏ด 39.963999 u ๏ฉ + 0.067302 ๏ด mass of 41K
๏จ
= 36.3368 u + 0.00468 u + 0.067302 ๏ด mass of 41K
mass of 41K =
48.
๏ฉ
๏ฉ
39.0983u ๏ญ ๏จ 36.3368 u + 0.00468 u ๏ฉ
= 40.962 u
0.067302
(M) We use the expression for determining the weighted-average atomic mass, where x
represents the fractional abundance of 10 B and ๏จ1 ๏ญ x ๏ฉ the fractional abundance of 11 B
10.81u = ๏จ10.012937 u ๏ด x ๏ฉ + ๏ฉ๏ซ11.009305 ๏ด ๏จ1 ๏ญ x ๏ฉ๏น๏ป = 10.012937 x +11.009305 ๏ญ 11.009305 x
10.81 ๏ญ 11.009305 = ๏ญ0.20 = 10.012937 x ๏ญ 11.009305 x = ๏ญ0.996368 x
0.20
= 0.20
x=
0.996368
๏ 20% 10 B and ๏จ100.0 ๏ญ 20๏ฉ = 80% 11B
Mass Spectrometry
49.
(M)
(a)
(b)
As before, we multiply each isotopic mass by its fractional abundance, after which,
we sum these products to obtain the (average) atomic mass for the element.
๏จ 0.205 ๏ด 70 ๏ฉ + ๏จ 0.274 ๏ด 72 ๏ฉ + ๏จ 0.078 ๏ด 73๏ฉ + ๏จ 0.365 ๏ด 74 ๏ฉ + ๏จ 0.078 ๏ด 76 ๏ฉ
= 14 + 20. + 5.7 + 27 + 5.9 = 72.6 = average atomic mass of germanium
The result is only approximately correct because the isotopic masses are given to only
two significant figures. Thus, only a two-significant-figure result can be quoted.
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Chapter 2: Atoms and the Atomic Theory
50.
(M)
(a) Six unique HCl molecules are possible (called isotopomers):
1 35
H Cl, 2 H 35Cl, 3 H 35Cl, 1H 37 Cl, 2 H 37 Cl, and 3 H 37 Cl
The mass numbers of the six different possible types of molecules are obtained by
summing the mass numbers of the two atoms in each molecule:
H 35 Cl has A ๏ฝ 36
1
H 37 Cl has A ๏ฝ 38
H 35 Cl has A ๏ฝ 37
2
H 37 Cl has A ๏ฝ 39
1
(b)
2
H 35 Cl has A ๏ฝ 38
3
H 37 Cl has A ๏ฝ 40
3
The most abundant molecule contains the
isotope for each element that is most abundant.
It is 1 H 35Cl . The second most abundant molecule
is 1 H 37 Cl . The relative abundance of each type of
molecule is determined by multiplying together the
fractional abundances of the two isotopes present.
Relative abundances of the molecules are as follows.
1
H 35 Cl : 75.76%
2
H 35 Cl : 0.011%
1
H 35 Cl : ๏ผ 0.0008%
3
3
H 37 Cl : 24.23%
2
H 37 Cl : 0.0036%
H 37 Cl : ๏ผ 0.0002%
The Periodic Table
51.
52.
(E)
(a)
Ge is in group 14 and in the fourth period.
(b)
Other elements in group 16(6A) are similar to S: O, Se, and Te. Most of the elements
in the periodic table are unlike S, but particularly metals such as Na, K, and Rb.
(c)
The alkali metal (group 1), in the fifth period is Rb.
(d)
The halogen (group 17) in the sixth period is At.
(E)
(a)
Au is in group 11 and in the sixth period.
(b)
Ar, Z = 18, is a noble gas. Xe is a noble gas with atomic number (54)
greater than 50.
(c)
If an element forms a stable anion with charge 2โ, it is in group 16.
(d)
If an element forms a stable cation with charge 3+, it is in group 13.
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Chapter 2: Atoms and the Atomic Theory
53.
(E) If the seventh period of the periodic table is 32 members long, it will be the same length
as the sixth period. Elements in the same family (vertical group), will have atomic numbers
32 units higher. The noble gas following radon will have atomic number = 86 + 32 = 118.
The alkali metal following francium will have atomic number = 87 + 32 = 119.
54.
(M) There are several interchanges: Ar/K, Co/Ni, Te/I, Th/Pa, U/Np, Pu/Am, Sg/Bh
The reverse order is necessary because the periodic table lists elements in order of increasing
atomic number (protons in the nucleus) and not in order of increasing atomic masses.
The Avogadro Constant and the Mole
55.
(E)
6.022 ๏ด 1023 atoms Fe
= 9.51 ๏ด 1024 atoms Fe
1 mol Fe
6.022 ๏ด 1023 atoms Ag
(b) atoms of Ag = 0.000467 mol Ag ๏ด
= 2.81 ๏ด 1020 atoms Ag
1 mol Ag
6.022 ๏ด 1023 atoms Na
= 5.1 ๏ด 1013 atoms Na
(c) atoms of Na = 8.5 ๏ด 10๏ญ11 mol Na ๏ด
1 mol Na
(a) atoms of Fe = 15.8 mol Fe ๏ด
56.
(E) Since the molar mass of nitrogen is 14.0 g/mol, 25.0 g N is almost two moles
(1.79 mol N), while 6.022 ๏ด1023 Ni atoms is about one mole, and 52.0 g Cr (52.00 g/mol
Cr) is also almost one mole. Finally, 10.0 cm 3 Fe (55.85 g/mol Fe) has a mass of about
79 g, and contains about 1.4 moles of atoms. Thus, 25.0 g N contains the greatest number
of atoms.
Note: Even if you take nitrogen as N2, the answer is the same.
57.
(E)
(a) moles of Zn = 415.0 g Zn ๏ด
1 mol Zn
= 6.347 mol Zn
65.38 g Zn
(b) # of Cr atoms = 147, 400 g Cr ๏ด
1 mol Cr
6.022 ๏ด 1023 atoms Cr
๏ด
51.996 g Cr
1 mol Cr
๏ฝ 1.707 ๏ด 1027 atoms Cr
1 mol Au
196.97 g Au
๏ด
๏ฝ 3.3 ๏ด 10 ๏ญ10 g Au
(c) mass Au ๏ฝ 1.0 ๏ด 1012 atoms Au ๏ด
23
6.022 ๏ด 10 atoms Au
1 mol Au
1 mol F
18.998 g F
(d) mass of F atom ๏ฝ 1 atom F ๏ด
๏ด
๏ฝ 3.1547 ๏ด 10๏ญ23 g F
23
6.022 ๏ด 10 atoms F
1 mol F
For exactly 1 F atom, the number of sig figs in the answer is determined by the least
precise number in the calculation, namely the mass of fluorine.
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Chapter 2: Atoms and the Atomic Theory
58.
(E)
(a) number Kr atoms = 5.25 mg Kr ๏ด
1 g Kr
1 mol Kr 6.022 ๏ด 1023 atoms Kr
๏ด
๏ด
1000 mg Kr 83.80 g Kr
1 mol Kr
= 3.77 ๏ด 1019 atoms Kr
(b) Molar mass is defined as the mass per mole of substance. mass = 2.09 g.
This calculation requires that the number of moles be determined.
1 mol
moles = 2.80 ๏ด1022 atoms ๏ด
= 0.0465 mol
6.022 ๏ด 1023 atoms
mass
2.09 g
molar mass =
=
= 44.9 g/mol The element is Sc, scandium.
moles 0.0465 mol
1 mol Mg
1 mol P 30.9738 g P
(c) mass P = 44.75 g Mg ๏ด
๏ด
๏ด
๏ฝ 57.03 g P
24.305 g Mg 1 mol Mg
1 mol P
Note: The same answer is obtained if you assume phosphorus is P4 instead of P.
59.
(E) Determine the mass of Cu in the jewelry, then convert to moles and finally to the
number of atoms. If sterling silver is 92.5% by mass Ag, it is 100 โ 92.5 = 7.5% by mass
Cu.
Conversion pathway approach:
number of Cu atoms ๏ฝ 33.24 g sterling ๏ด
7.5 g Cu
1 mol Cu
6.022 ๏ด 1023 atoms Cu
๏ด
๏ด
100.0 g sterling 63.546 g Cu
1 mol Cu
๏ฝ 2.4 ๏ด 1022 Cu atoms
Stepwise approach:
7.5 g Cu
๏ฝ 2.493 g Cu
100.0 g sterling
1 mol Cu
๏ฝ 0.03923 mol Cu
2.493 g Cu ๏ด
63.546 g Cu
33.24 g sterling ๏ด
0.03923 mol Cu ๏ด
60.
6.022 ๏ด 1023 atoms Cu
= 2.4 ๏ด 1022 Cu atoms
1 mol Cu
(E) We first need to determine the amount in moles of each metal.
9.4 g solder
67 g Pb
1 mol Pb
amount of Pb = 50.0 cm3 solder ๏ด
๏ด
๏ด
= 1.5 mol Pb
3
1 cm
100 g solder 207.2 g Pb
9.4 g solder
33 g Sn
1 mol Sn
amount of Sn = 50.0 cm3 solder ๏ด
๏ด
๏ด
= 1.3 mol Sn
3
1 cm
100 g solder 118.7 g Sn
total atoms = ๏จ1.5 mol Pb +1.3 mol Sn ๏ฉ ๏ด
6.022 ๏ด 1023 atoms
= 1.7 ๏ด 1024 atoms
1 mol
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Chapter 2: Atoms and the Atomic Theory
61.
(E) We first need to determine the number of Pb atoms of all types in 215 mg of Pb, and
then use the percent isotopic abundance to determine the number of 204 Pb atoms present.
1g
1 mol Pb 6.022 ๏ด 1023atoms 14 204 Pb atoms
204
Pb atoms ๏ฝ 215 mg Pb ๏ด
๏ด
๏ด
๏ด
1000 mg 207.2 g Pb
1 mol Pb
1000 Pb atoms
๏ฝ 8.7 ๏ด 1018 atoms 204 Pb
62.
(E)
mass of alloy = 7.25 ๏ด 1023 Cd atoms ๏ด
1 mol Cd
112.4 g Cd 100.0 g alloy
๏ด
๏ด
23
6.022 ๏ด 10 Cd atoms 1 mol Cd
8.0 g Cd
= 1.7 ๏ด 103 g alloy
63.
(E) We will use the average atomic mass of lead, 207.2 g/mol, to answer this question.
30 ๏ญg Pb 1 dL
1 g Pb
1 mol Pb
(a)
๏ด
๏ด 6
๏ด
๏ฝ 1.45 ๏ด 10๏ญ6 mol Pb / L
1 dL
0.1 L 10 ๏ญg Pb 207.2 g
1.45 ๏ด 10๏ญ6 mol Pb
1L
6.022 ๏ด 1023 atoms
(b)
๏ด
๏ด
๏ฝ 8.7 ๏ด 1014 Pb atoms / mL
L
1000 mL
1 mol
(M) The concentration of Pb in air provides the principal conversion factor. Other
conversion factors are needed to convert to and from its units, beginning with the 0.500 L
volume, and ending with the number of atoms.
1 m3
3.11 ๏ญg Pb
1 g Pb
1 mol Pb 6.022 ๏ด 1023 Pb atoms
no. Pb atoms ๏ฝ 0.500 L ๏ด
๏ด
๏ด 6
๏ด
๏ด
1000 L
1 m3
10 ๏ญg Pb 207.2 g Pb
1 mol Pb
64.
๏ฝ 4.52 ๏ด 1012 Pb atoms
65.
(M) To answer this question, we simply need to calculate the ratio of the mass (in grams)
of each sample to its molar mass. Whichever elemental sample gives the largest ratio will be
the one that has the greatest number of atoms.
(a) Iron sample: 10 cm ๏ด 10 cm ๏ด 10 cm ๏ด 7.86 g cmโ3 = 7860 g Fe
1 mol Fe
7860 g Fe ๏ด
= 141 mol of Fe atoms
55.845 g Fe
1.00 ๏ด 103g H 2
(b) Hydrogen sample:
๏ด 1 mol H = 496 mol of H2 molecules =
2 ๏ด (1.008 g H)
992 mol of H atoms
4
2.00 ๏ด 10 g S
(c) Sulfur sample:
๏ด 1 mol S = 624 mol of S atoms
32.06 g S
454 g Hg
1 mol Hg
๏ด
= 172 mol of Hg atoms
(d) Mercury sample: 76 lb Hg ๏ด
1 lb Hg
200.6 g Hg
Clearly, then, it is the 1.00 kg sample of hydrogen that contains the greatest number of atoms.
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Chapter 2: Atoms and the Atomic Theory
66.
(M)
(a) 23 g Na = 1 mol with a density ~ 1 g/cm3. 1 mole = 23 g, so volume of 25.5 mol ~ 600 cm3.
(b) Liquid bromine occupies 725 mL or 725 cm3 (given).
(c) 1.25 ร 1025 atoms Cr is ~20 moles. At ~50 g/mol, this represents approximately 1000 g.
Given the density of 9.4 g/cm3, this represents about 100 cm3 of volume.
(d) 2150 g solder at 9.4 g/cm3 represents approximately 200 cm3.
From this we can see that the liquid bromine would occupy the largest volume.
INTEGRATIVE AND ADVANCED EXERCISES
67. (M)
0.9922 g
= 2.50 g + 99.22 g =101.72 g
1mL
1.0085g
mass of solution (20 ๏ฐC) = 100 mL ร
= 100.85g
1mL
solid crystallized = total mass(40 ๏ฐC) โ solution mass(20๏ฐC)
= 101.72 g โ 100.85 g = 0.87 g
(a) total mass(40 ๏ฐC) = 2.50 g + 100.0 mL ร
(b) The answer cannot be more precise because both the initial mass and the subtraction only
allows the reporting of masses to + 0.01 g. For a more precise answer, more significant
figures would be required for the initial mass (2.50 g) and the densities of the water and
solution.
68. (M) We now recognize that the values of 24.3 u and 35.3 u for the masses of Mg and Cl
represent a weighted-average that considers the mass and abundance of each isotope. The
experimental mass of each isotope is very close to the natural number and therefore very
close to an integer multiple of the mass of 1H, thus supporting Proutโs hypothesis.
69. (M) Each atom of 19 F contains 9 protons (1.0073 u each), 10 neutrons (1.0087 u each), and
9 electrons (0.0005486 u each). The mass of each atom should be the sum of the masses of
these particles.
๏ฆ
1.0073 u ๏ถ ๏ฆ
1.0087 u ๏ถ ๏ฆ
0.0005486 u ๏ถ
Total mass ๏ฝ ๏ง 9 protons ๏ด
๏ซ ๏ง10 neutrons ๏ด
๏ซ ๏ง 9 electrons ๏ด
๏ท
๏ท
1 proton ๏ธ ๏จ
1 neutron ๏ธ ๏จ
1 electron ๏ท๏ธ
๏จ
= 9.0657 u + 10.087 u + 0.004937 u = 19.158 u
This compares with a mass of 18.9984 u given in the periodic table. The difference,
0.160 u per atom, is called the mass defect and represents the energy that holds the nucleus
together, the nuclear binding energy. This binding energy is released when 9 protons and
9 neutrons fuse to give a fluorine-19 nucleus.
70. (M)
4
volume of nucleus(single proton) ๏ฝ ๏ฐ r 3 ๏ฝ 1.3333 ๏ด 3.14159 ๏ด (0.5 ๏ด 10 ๏ญ13 cm)3 ๏ฝ 5 ๏ด 10 ๏ญ40 cm3
3
๏ญ24
1.673 ๏ด 10 g
density ๏ฝ
๏ฝ 3 ๏ด 1015 g/cm 3
๏ญ 40
3
5 ๏ด 10 cm
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Chapter 2: Atoms and the Atomic Theory
71. (M) This method of establishing the Avogadro constant uses fundamental constants.
1 mol 12 C 6.022 ๏ด 1023 12 C atoms
12 u
12
๏ด
๏ด 12
๏ฝ 6.022 ๏ด 1023 u
1.000 g C ๏ด
12
12
12 g C
1 mol C
1 C atom
72. (M) Let Z = # of protons, N = # of neutrons, E = # of electrons, and
A = # of nucleons = Z + N.
(a) Z + N = 234 The mass number is 234 and the species is an atom.
N = 1.600 Z The atom has 60.0% more neutrons than protons.
Next we will substitute the second expression into the first and solve for Z.
Z ๏ซ N ๏ฝ 234 ๏ฝ Z ๏ซ 1.600 Z ๏ฝ 2.600 Z
234
๏ฝ 90 protons
Z๏ฝ
2.600
Thus this is an atom of the isotope 234 Th.
(b) Z = E + 2
The ion has a charge of +2. Z = 1.100 E
There are 10.0% more protons than electrons. By equating these two expressions and
solving for E, we can find the number of electrons. E ๏ซ 2 ๏ฝ 1.100 E
2
E๏ฝ
๏ฝ 20 electrons
Z ๏ฝ 20 ๏ซ 2 ๏ฝ 22, (titanium).
2 ๏ฝ 1.100 E ๏ญ E ๏ฝ 0.100 E
0.100
The ion is Ti 2 ๏ซ . There is not enough information to determine the mass number.
(c) Z + N = 110 The mass number is 110.
Z=E+2
The species is a cation with a charge of +2.
N = 1.25 E
Thus, there are 25.0% more neutrons than electrons. By substituting the
second and third expressions into the first, we can solve for E, the number of electrons.
108
E๏ฝ
๏ฝ 48
( E ๏ซ 2) ๏ซ 1.25 E ๏ฝ 110 ๏ฝ 2.25 E ๏ซ 2 108 ๏ฝ 2.25 E
2.25
Then Z ๏ฝ 48 ๏ซ 2 ๏ฝ 50, (the element is Sn) N ๏ฝ 1.25 ๏ด 48 ๏ฝ 60 Thus, it is 110Sn 2๏ซ .
73. (E) Because the net ionic charge (+ 2) is one-tenth of its the nuclear charge, the nuclear charge
is + 20. This is also the atomic number of the nuclide, which means the element is calcium.
The number of electrons is 20 for a neutral calcium atom, but only 18 for a calcium ion with a
2๏ซ
net 2+ charge. Four more than 18 is 22 neutrons. The ion described is 42
20 Ca .
74. (M) A = Z + N = 2.50 Z The mass number is 2.50 times the atomic number.
The neutron number of selenium-82 equals 82 โ 34 = 48, since Z = 34 for Se. The neutron
number of isotope Y also equals 48, which equals 1.33 times the atomic number of isotope Y.
48
๏ฝ 36
Thus 48 ๏ฝ 1.33 ๏ด Z Y Z Y ๏ฝ
1.33
The mass number of isotope Y = 48 + 36 = 84 = the atomic number of E, and thus, the
element is Po. Thus, from the relationship in the first line, the mass number of
E ๏ฝ 2.50 Z ๏ฝ 2.50 ๏ด 84 ๏ฝ 210
The isotope E is 210 Po .
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Chapter 2: Atoms and the Atomic Theory
75. As a result of the redefinition, all masses will decrease by a factor of
35.00000/35.453 = 0.98722.
(a) atomic mass of He
atomic mass of Na
atomic mass of I
4.00260 ๏ด 0.98722 ๏ฝ 3.9515
22.9898 ๏ด 0.98722 ๏ฝ 22.696
126.905 ๏ด 0.98722 ๏ฝ 125.28
(b) These three elements have nearly integral atomic masses based on C-12 because these
three elements and C-12 all consist mainly of one stable isotope, rather than a mixture
of two or more stable isotopes, with each being present in significant amounts (10% or
more), as is the case with chlorine.
76. (M) To solve this question, represent the fractional abundance of 14 N by x and that of 14 N
by (1 โ x). Then use the expression for determining average atomic mass.
14.0067 ๏ฝ 14.0031 x ๏ซ 15.0001(1 ๏ญ x)
14.0067 ๏ญ 15.0001 ๏ฝ 14.0031 x ๏ญ 15.0001 x
OR
๏ญ0.9934 ๏ฝ ๏ญ0.9970 x
0.9934
๏ด 100% ๏ฝ 99.64% ๏ฝ percent isotopic abundance of 14 N.
0.9970
Thus, 0.36% ๏ฝ percent isotopic abundance of 15 N.
x๏ฝ
77. (D) In this case, we will use the expression for determining average atomic massโthe sum of
products of nuclidic mass times fractional abundances (from Figure 2-14)โto answer the
question.
196
Hg: 195.9658 u ๏ด 0.00146 ๏ฝ 0.286 u
198
199
197.9668 u ๏ด 0.1002 ๏ฝ 19.84 u
198.9683 u ๏ด 0.1684 ๏ฝ 33.51 u
Hg :
Hg :
200
201
200.9703 u ๏ด 0.1322 ๏ฝ 26.57 u
Hg : 199.9683 u ๏ด 0.2313 ๏ฝ 46.25 u
Hg :
202
204
201.9706 u ๏ด 0.2980 ๏ฝ 60.19 u
203.9735 u ๏ด 0.0685 ๏ฝ 14.0 u
Hg :
Hg :
Atomic weight = 0.286 u + 19.84 u + 33.51 u + 46.25 u + 26.57 u + 60.19 u + 14.0 u = 200.6 u
78. (D) The sum of the percent abundances of the two minor isotopes equals 100.00% โ 84.68% =
15.32%. Thus, we denote the fractional abundance of 73Ge as x, and the other as (0.1532 โ x).
These fractions are then used in the expression for average atomic mass.
atomic mass ๏ฝ 72.64 u ๏ฝ (69.92425 u ๏ด 0.2085) ๏ซ (71.92208 u ๏ด 0.2754) ๏ซ (73.92118 u ๏ด 0.3629)
๏ซ (72.92346 u ๏ด x) + [75.92140 u ๏ด (0.1532 ๏ญ x)]
72.64 u ๏ฝ (14.5792 u) ๏ซ (19.8073 u) ๏ซ (26.8260 u) ๏ซ (72.92346 u x) ๏ซ [11.6312 u ๏ญ 75.92140 u x)]
๏ญ0.2037 ๏ฝ ๏ญ2.99794x Hence: x ๏ฝ 0.068.
73
Ge has 6.8% isotopic abundance and 76Ge has 8.5% isotopic abundance.
From the calculations, we can see that the number of significant figures drops from four
in the percent isotopic abundances supplied to only two significant figures owing to the
imprecision in the supplied values of the percent isotopic abundance.
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Chapter 2: Atoms and the Atomic Theory
79. (D) First, it must be understood that because we donโt know the exact percent abundance of
84
Kr, all the percent abundances for the other isotopes will also be approximate. From the
question, we may initially infer the following:
๏ท Assume percent abundance of 84Kr ~ 55% as a start (somewhat more than 50)
๏ท Let percent abundance of 82Kr = x%; percent abundance 83Kr ~ 82Kr = x%
๏ท 86Kr = 1.50(percent abundance of 82Kr) = 1.50(x%)
๏ท 80Kr = 0.196(percent abundance of 82Kr) = 0.196(x%)
๏ท 78Kr = 0.030(percent abundance of 82Kr) = 0.030(x%)
100% = %78Kr + %80Kr + %82Kr + %83Kr + %84Kr + %86Kr
100% = 0.030(x%) + 0.196(x%) + x% + x% + %84Kr + 1.50(x%)
100% = 3.726(x%) + %84Kr
Assuming percent abundance of 84Kr is 55%, solving for x gives a value of 12.1% for percent
abundance of 82Kr, from which the remaining abundances can be calculated based on the
above relationships, as shown below:
78
Kr: 0.03 ร 12.1 = 0.363%; 80Kr: 0.196 ร 12.1 = 2.37%; 83Kr: same as 82Kr; 86Kr: 1.5 ร 12.1
= 18.15%.
The weighted-average isotopic mass calculated from the above abundances is as follows:
Weighted-average isotopic mass = 0.030(12.1%)(77.9204 u) + 0.196(12.1%)(79.9164 u) +
12.1%(81.9135 u) + 12.1%(82.9141 u) + 55%(83.9115 u) + 1.50(12.1%)(85.9106 u) =
83.8064 u
As stated above, the problem here is the inaccuracy of the percent abundance for 84Kr, which
is crudely estimated to be ~ 55%. If we vary this percentage, we vary the relative abundance
of all other isotopes accordingly. Since we know the weighted-average atomic mass of Kr is
83.80, we can try different values for 84Kr abundance and figure out which gives us the
closest value to the given weighted-average isotopic mass:
Percent
Weighted-Average
Abundance 84Kr
Isotopic Mass
50%
83.793
51%
83.796
52%
83.799
53%
83.801
54%
83.803
55%
83.806
From this table, we can see that the answer is somewhere between 52% and 53%.
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Chapter 2: Atoms and the Atomic Theory
80. (D) Four molecules are possible, given below with their calculated molecular masses.
35
Cl 79Br mass = 34.9689 u + 78.9183 u = 113.8872 u
35
Cl 81Br mass = 34.9689 u + 80.9163 u = 115.8852 u
37
Cl 79Br mass = 36.9658 u + 78.9183 u = 115.8841 u
37
Cl 81Br mass = 36.9658 u + 80.9163 u = 117.8821 u
Each molecule has a different intensity pattern (relative number of molecules), based on the
isotopic abundance of the isotopes making up each molecule. If we divide all of the values by
the lowest ratio, we can get a better idea of the relative ratio of each molecule.
35
Cl 79Br Intensity = (0.7577) ร (0.5069 ) = 0.3841 ๏ธ 0.1195 = 3.214
35
Cl 81Br Intensity = (0.7577) ร (0.4931 ) = 0.3736 ๏ธ 0.1195 = 3.127
37
Cl 79Br Intensity = (0.2423) ร (0.5069 ) = 0.1228 ๏ธ 0.1195 = 1.028
37
Cl 81Br Intensity = (0.2423) ร (0.4931 ) = 0.1195 ๏ธ 0.1195 = 1.000
A plot of intensity versus molecular mass reveals the following pattern under ideal
circumstances (high resolution mass spectrometry).
81. (M) Letโs begin by finding the volume of copper metal.
2.54 cm
= 0.08118 cm
wire diameter (cm) = 0.03196 in. ๏ด
1 in.
The radius is 0.08118 cm ๏ด 1/2 = 0.04059 cm
The volume of Cu(cm3) = ๏ฐ r2 h = 3.1416 ๏ด (0.04059 cm)2 ๏ด 150 cm = 0.7764 cm3
8.92 g Cu
So, the mass of Cu = 0 cm3 ๏ด
= 6.93 g Cu
1 cm3
1 mol Cu
The number of moles of Cu = 6.93 g Cu ๏ด
= 0.109 mol Cu
63.546 g Cu
6.022 ๏ด 1023 atoms Cu
Cu atoms in the wire = 0.109 mol Cu ๏ด
= 6.56 ๏ด 1022 atoms
1 mol Cu
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Chapter 2: Atoms and the Atomic Theory
82. (D) volume = l ยด wยด h – p r 2 h
2
รฆ 2.50 cm รทรถ
= (15.0 cm ยด12.5 cm ยด 0.300 cm) – (3.1416) ยดรงรง
ยด(0.300 cm)
รงรจ 2 รทรทรธ
= (56.25 cm -1.47 cm) = 54.8 cm3
mass of object = 54.8 cm3 ยด
8.80 g
= 482 g Monel metal
1 cm3
Then determine the number of silicon atoms in this quantity of alloy.
2.2ยด10-4 g Si 1 mol Si 6.022ยด1023 Si atoms
482 g Monel metalยด
ยด
ยด
= 2.3ยด1021 Si atoms
1.000 g metal 28.05 g Si
1 mol Si
Finally, determine the number of 30 Si atoms in this quantity of silicon.
รฆ 3.10 30 Si atoms รถรท
รท = 7.1ยด1019 30 Si
number of Si atoms = (2.3ยด10 Si atoms)ยดรงรง
รงรจ 100 Si atoms รธรทรท
30
83.
21
(M) The percent isotopic abundance of deuterium of 0.015% means that, in a sample of
100,000 H atoms, only 15 2H are present.
mass H 2 = 2.50ร10 21 atoms 2 H ร
100,000 H atoms
2
15 atoms H
ร
1 mol H
23
6.022ร10 H atoms
ร
1 mol H 2
2 mol H
ร
2.016 g H 2
1 mol H 2
mass H 2 = 27.9 g H 2 is required such that we have 2.50ร1021 atoms 2 H.
84. (M) The numbers sum to 21 (= 10 + 6 + 5). Thus, in one mole of the alloy there is 10 21 mol
Bi, 6 21 mol Pb, and 5 21 mol Sn. The mass of this mole of material is figured in a similar
fashion to computing a weighted-average atomic mass from isotopic masses.
๏ฆ 10
209.0 g ๏ถ ๏ฆ 6
207.2 g ๏ถ ๏ฆ 5
118.7 ๏ถ
๏ท๏ท ๏ซ ๏ง๏ง mol Pb ๏ด
๏ท๏ท ๏ซ ๏ง๏ง mol Sn ๏ด
๏ท
mass of alloy ๏ฝ ๏ง๏ง mol Bi ๏ด
1 mol Bi ๏ธ ๏จ 21
1 mol Pb ๏ธ ๏จ 21
1 mol Sn ๏ท๏ธ
๏จ 21
๏ฝ 99.52 g Bi ๏ซ 59.20 g Pb ๏ซ 28.26 g Sn ๏ฝ 186.98 g alloy
85. (M) The atom ratios are of course, the same as the mole ratios. We first determine the mass of
alloy that contains 5.00 mol Ag, 4.00 mol Cu, and 1.00 mol Zn.
107.87 g Ag
63.546 g Cu
65.38g Zn
๏ซ 4.00 mol Cu ๏ด
๏ซ 1.00 mol Zn ๏ด
1mol Ag
1mol Cu
1mol Zn
๏ฝ 539.4 g Ag ๏ซ 254.2 g Cu ๏ซ 65.38g Zn ๏ฝ 859 g alloy
mass ๏ฝ 5.00 mol Ag ๏ด
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Chapter 2: Atoms and the Atomic Theory
Then, for 1.00 kg of the alloy, (1000 g), we need 1000/859 moles. (859 g is the alloyโs
โmolar mass.โ)
539 g Ag
mass of Ag ๏ฝ 1000 g alloy ๏ด
๏ฝ 627 g Ag
859 g alloy
254 g Cu
mass of Cu ๏ฝ 1000 g alloy ๏ด
๏ฝ 296 g Cu
859 g alloy
65.4 g Zn
mass of Zn ๏ฝ 1000 g alloy ๏ด
๏ฝ 76.1 g Zn
859 g alloy
86. (M) The relative masses of Sn and Pb are 207.2 g Pb (assume one mole of Pb) to
(2.73 ร 118.71 g/mol Sn =) 324 g Sn. Then the mass of cadmium, on the same scale,
is 207.2/1.78 = 116 g Cd.
324 g Sn
324 g Sn
% Sn ๏ฝ
๏ด 100% ๏ฝ
๏ด 100% ๏ฝ 50.1%Sn
207.2 ๏ซ 324 ๏ซ 116 g alloy
647 g alloy
% Pb ๏ฝ
207.2 g Pb
๏ด 100% ๏ฝ 32.0% Pb
647 g alloy
% Cd ๏ฝ
116 g Cd
๏ด 100% ๏ฝ 17.9% Cd
647 g alloy
87. (M) We need to apply the law of conservation of mass and convert volumes to masses:
Calculate the mass of zinc:
125 cm3 ร 7.13 g/cm3 = 891 g
Calculate the mass of iodine:
125 cm3 ร 4.93 g/cm3 = 616 g
Calculate the mass of zinc iodide:
164 cm3 ร 4.74 g/cm3 = 777 g
(891 + 616 โ 777) g = 730 g
Calculate the mass of zinc unreacted:
Calculate the volume of zinc unreacted:
730 g ร 1cm3 / 7.13 g = 102 cm3
88. (D) First, calculate the total number of Ag atoms in a 1 cm3 crystal:.
1 mol Ag
6.02 ๏ด 1023 Ag atoms
10.5 g Ag ๏ด
๏ด
๏ฝ 5.86 ๏ด 1022 atoms of Ag
107.87 g Ag
1 mol Ag
The actual volume taken up by the Ag atoms (considering that there is 26% empty space in
the crystal) is: 1 cm3 ร 0.74 = 0.74 cm3
Therefore, the volume of each atom is:
0.74 cm3
= 1.263 ร 10โ23 cm3
5.86 ๏ด 1022 atoms
Volume of a sphere is expressed as V = 1.263 ร 10โ23 cm3 = (4/3) ฯ r3
Solving for r, we get 1.44 ร 10โ8 cm or 144 pm.
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Chapter 2: Atoms and the Atomic Theory
FEATURE PROBLEMS
89.
(M) The product mass differs from that of the reactants by ๏จ 5.62 ๏ญ 2.50 = ๏ฉ 3.12 grains. In
order to determine the percent gain in mass, we need to convert the reactant mass to grains.
8 gros
72 grains
13 onces ๏ด
= 104 gros ๏ด ๏จ104 + 2 ๏ฉ gros ๏ด
= 7632 grains
1 once
1 gros
3.12 grains increase
% mass increase ๏ฝ
๏ด 100% ๏ฝ 0.0409% mass increase
๏จ7632 + 2.50๏ฉ grains original
The sensitivity of Lavoisierโs balance can be as little as 0.01 grain, which seems to be the
limit of the readability of the balance; alternatively, it can be as large as 3.12 grains, which
assumes that all of the error in the experiment is due to the (in)sensitivity of the balance. Let
us convert 0.01 grains to a mass in grams.
1 gros 1 once 1 livre 30.59 g
๏ด
๏ด
๏ด
๏ฝ 3 ๏ด 10๏ญ5 g ๏ฝ 0.03 mg
minimum error ๏ฝ 0.01 gr ๏ด
72 gr 8 gros 16 once 1 livre
3 ๏ด 10 ๏ญ5 g
maximum error ๏ฝ 3.12 gr ๏ด
๏ฝ 9 ๏ด 10 ๏ญ3 g ๏ฝ 9 mg
0.01 gr
The maximum error is close to that of a common modern laboratory balance, which has a
sensitivity of 1 mg. The minimum error is approximated by a good quality analytical
balance. Thus we conclude that Lavoisierโs results conform closely to the law of
conservation of mass.
90.
(D) One way to determine the common factor of which all 13 numbers are multiples is to
first divide all of them by the smallest number in the set. The ratios thus obtained may be
either integers or rational numbers whose decimal equivalents are easy to recognize.
1
2
3
4
5
6
7
8
9
10
11
12
13
Obs.
Quan. 19.6 24.60 29.62 34.47 39.38 44.42 49.41 53.91 59.12 63.68 68.65 78.34 83.22
Ratio 1.00 1.251 1.507 1.753 2.003 2.259 2.513 2.742 3.007 3.239 3.492 3.984 4.233
Mult. 4.00 5.005 6.026 7.013 8.012 9.038 10.05 10.97 12.03 12.96 13.97 15.94 16.93
Int.
4
5
6
7
8
9
10
11
12
13
14
16
17
The row labeled โMult.โ is obtained by multiplying the row โRatioโ by 4.000. In the row
labeled โInt.โ we give the integer closest to each of these multipliers. It is obvious that each
of the 13 measurements is exceedingly close to a common quantity multiplied by an integer.
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Chapter 2: Atoms and the Atomic Theory
91.
(M) In a 60-year-old chemistry textbook, the atomic mass for oxygen would be exactly
16 u because chemists assigned precisely 16 u as the atomic mass of the naturally occurring
mixture of oxygen isotopes. This value is slightly higher than the value of 15.9994 in
modern chemistry textbooks. Thus, we would expect all other atomic masses to be slightly
higher as well in the older textbooks.
92.
(D) We begin with the amount of reparations and obtain the volume in cubic kilometers
with a series of conversion factors.
Conversion pathway approach:
1troy oz Au 31.103g Au
1mol Au
6.022 ๏ด 1023 atoms Au
9
V ๏ฝ $28.8 ๏ด 10 ๏ด
๏ด
๏ด
๏ด
$21.25
1troy oz Au 196.97 g Au
1mol Au
1ton seawater
2000 lb seawater 453.6 g seawater 1cm 3 seawater
๏ด
๏ด
๏ด
๏ด
4.67 ๏ด 1017 Au atoms
1ton seawater
1 lb seawater
1.03g seawater
3
๏ฆ 1m
1km ๏ถ
๏ด๏ง
๏ด
๏ฝ 2.43 ๏ด 105 km3
๏ท
๏จ 100 cm 1000 m ๏ธ
Stepwise approach:
$28.8 ๏ด109 ๏ด
1troy oz Au 31.103g Au
๏ด
๏ฝ 4.22 ๏ด 1010g Au
$21.25
1troy oz Au
4.22 ๏ด1010 g Au ๏ด
1mol Au
6.022 ๏ด 1023 atoms Au
๏ด
๏ฝ 1.29 ๏ด 1032 atoms Au
196.97 g Au
1mol Au
1.29 ๏ด1032 atoms Au ๏ด
1ton seawater
2000 lb seawater
๏ฝ 5.52 ๏ด1017 lb seawater
๏ด
17
4.67 ๏ด10 Au atoms 1ton seawater
5.52 ๏ด1017 lb seawater ๏ด
453.6 g seawater 1cm 3 seawater
๏ด
๏ฝ 2.43 ๏ด1020cm3 seawater
1 lb seawater
1.03g seawater
3
๏ฆ 1m
1km ๏ถ
5
3
๏ด
2.43 ๏ด10 cm seawater ๏ด ๏ง
๏ท ๏ฝ 2.43 ๏ด10 km
100
cm
1000
m
๏จ
๏ธ
20
3
93. (D) We start by using the percent isotopic abundances for 87Rb and 85Rb along with the data
in the โspikedโ mass spectrum to find the total mass of Rb in the sample. Then, we calculate
the Rb content in the rock sample in ppm by mass by dividing the mass of Rb by the total
mass of the rock sample, and then multiplying the result by 106 to convert to ppm.
87
Rb = 27.83% isotopic abundance 85Rb = 72.17% isotopic abundance
87
Rb(isotopic)
27.83%
=
= 0.3856
Therefore, 85
Rb(isotopic)
72.17%
For the 87Rb(spiked) sample, the 87Rb peak in the mass spectrum is 1.12 times as tall as the
87
Rb(isotopic) ๏ซ 87 Rb(spiked)
85
Rb peak. Thus, for this sample
= 1.12
85
Rb(isotopic)
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Chapter 2: Atoms and the Atomic Theory
Using this relationship, we can now find the masses of both 85Rb and 87Rb in the sample.
87
87
Rb(isotopic)
Rb(isotopic)
85
So,
= 0.3856;
Rb(isotopic) =
85
Rb(isotopic)
0.3856
1.12 ๏ด 87 Rb (isotopic)
= 2.905 87Rb(isotopic)
0.3856
87
Rb(spiked) = 1.905 87Rb(isotopic)
87
87
Rb(isotopic) ๏ซ 87 Rb(spiked)
Rb(isotopic) ๏ซ 87 Rb(spiked)
=
and
= 1.12
87
85
Rb(isotopic)
Rb(isotopic)
0.3856
Since the mass of 87Rb(spiked) is equal to 29.45 ๏ญg, the mass of 87Rb(isotopic) must be
29.45 ๏ญg
= 15.46 ๏ญg of 87Rb(isotopic)
1.905
15.46 ๏ญg of 87 Rb(isotopic)
= 40.09 ๏ญg of 85Rb(isotopic)
So, the mass of 85Rb(isotopic) =
0.3856
Therefore, the total mass of Rb in the sample = 15.46 ๏ญg of 87Rb(isotopic) + 40.09 ๏ญg of
85
Rb(isotopic) = 55.55 ๏ญg of Rb. Convert to grams:
1 g Rb
= 5.555 ๏ด 10โ5 g Rb
= 55.55 ๏ญg of Rb ๏ด
1 ๏ด 106 ฮผg Rb
87
Rb(isotopic) + 87Rb(spiked) =
Rb content (ppm) =
5.555 ๏ด 10 ๏ญ5 g Rb
๏ด 106 = 159 ppm Rb
0.350 g of rock
SELF-ASSESSMENT EXERCISES
94.
(E)
(a)
A
Z
E : The element โEโ with the atomic number of Z (i.e., Z protons in the nucleus) and
atomic mass of A (i.e., total of protons and neutrons equals A).
(b) ฮฒ particle: An electron produced as a result of the decay of a neutron
(c)
Isotope: Nuclei that share the same atomic number but have different atomic masses
(d)
16
(e)
Molar mass: Mass of one mole of a substance
O: An oxygen nucleus containing 8 neutrons
56
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Chapter 2: Atoms and the Atomic Theory
95.
(E)
(a) The total mass of substances present after the chemical reaction is the same as the total
mass of substances before the chemical reaction. More universally, mass is neither
created nor destroyed, but converts from one form to another.
(b) Rutherfordโs model of the atom postulates the existence of positively charged
fundamental particles at the nucleus of the atom.
(c) An average value used to express the atomic mass of an element by taking into account
the atomic masses and relative abundances of all the naturally occurring isotopes of the
element.
(d) A spectrum showing the mass/charge ratio of various atoms in a matrix
96.
(E)
(a) Cathode rays are beams of electrons being generated from a negatively charged surface
(cathode) moving to a positively charged surface (anode). X-rays are high energy
photons, which are typically generated when the high energy beam of electrons
impinges on the anode.
(b) Protons and neutrons are both fundamental particles that make up an atomโs nucleus.
Protons are positively charged and neutrons have no charge.
(c) Nuclear charge is determined by the numbers of protons in the nucleus. Ionic charge is
determined by the difference between the number of protons and electrons in the atom.
(d) Periods are horizontal rows in the periodic table, while groups are vertical columns.
(e) Metals are generally characterized by their malleability, ductility, and ability to
conduct electricity and heat well. Nonmetals are generally brittle and nonconductive.
(f) The Avogadro constant is the number of elementary entities (atoms, molecules, etc.) in
one mole of a substance. A mole is the amount of a substance that contains the same
number of elementary entities as there are atoms in exactly 12 g of pure carbon-12.
97.
๏ฆ1
๏ถ ๏ฆ4
๏ถ
(E) ๏ง ร 10.013 u ๏ท ๏ซ ๏ง ๏ด 11.009 u ๏ท ๏ฝ 10.810 u. Therefore, the element is boron.
๏จ5
๏ธ ๏จ5
๏ธ
98.
(E) The answer is (b). If all of the zinc reacts and the total amount of the product (zinc
sulfide) is 14.9 g, then 4.9 g of S must have reacted with zinc. Therefore, 3.1 g of S remain.
99.
(E) The answer is (d). It should be remembered that atoms combine in ratios of whole
numbers. Therefore:
(a) 16 g O ร (1 mol O/16 g O) = 1 mol O, and 85.5 g Rb ร (1 mol Rb/85.5 g Rb) = 1 mol Rb
Therefore, the O: Rb ratio is 1:1.
(b) Same calculations as above give an O: Rb ratio of 0.5:0.5, or 1:1.
(c)
Same type calculation gives an O: Rb ratio of 2:1.
Because all of the above combine in O and Rb in whole number ratios, they are all at least
theoretically possible.
57
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Chapter 2: Atoms and the Atomic Theory
100. (E) If the compound contains 46.7% X, then the percent of oxygen would be 100.0% โ
46.7 = 53.3%. Assuming the sample is 100.0 g, there would be 53.3 g of oxygen present.
๏ฆ 1 mol O ๏ถ
53.3 g O ๏ง
๏ท ๏ฝ 3.331 mol O
๏จ 16.00 g O ๏ธ
The formula of the compound is XO; therefore, X and O are in a 1:1 mole ratio.
46.7 g X
๏ฝ 14.0 g/mol X
3.331 mol X
101. (E) The answer is (c). Cathode rays are beams of electrons, and as such have identical
properties to ฮฒ particles, although they may not have the same energy.
102. (E) The answer is (a), that the greatest portion of the mass of an atom is concentrated in a
small but positively charged nucleus.
103. (E) The answer is (d). A hydrogen atom has one proton and one electron, so its charge is
zero. A neutron has the same charge as a proton, but is neutral. Since most of the mass of the
atom is at the nucleus, a neutron has nearly the same mass as a hydrogen atom.
104. (E) All the choices in this question are fundamental particles (e).
105. (E) Dalton (d) is correct.
106. (E) There are no particles that have the same mass as the hydrogen atom and a negative
charge. The correct answer is (e).
35
107. (E) 17
Cl๏ซ
108. (E) The answer is (d), calcium, because they are in the same group.
109. (E) (a) Group 18, (b) Group 17, (c) Group 13 and Group 1 (d) Group 18
110. (E) (d) and (f)
111. (E) (c), because it is not close to being a whole number
112. (M) The answer is (d). Even with the mass scale being redefined based on 84Xe, the mass
ratio between 12C and 84Xe will remain the same. Using 12C as the original mass scale, the
mass ratio of 12C : 84Xe is 12 u/83.9115 u = 0.1430. Therefore, redefining the mass scale by
assigning the exact mass of 84 u to 84Xe, the relative mass of 12C becomes 84 ร 0.14301 =
12.0127 u.
58
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Chapter 2: Atoms and the Atomic Theory
113. (M) The answer is (b)
1000 g 1 mol Fe
๏ด
๏ฝ 1.000 ๏ด 10 ๏ญ2 mol Fe
5.585 kg Fe ๏ด
1 kg
55.85 g Fe
1 mol C
๏ฝ 50.01 mol C
600.6 g C ๏ด
12.01 g C
Therefore, 100 moles of Fe has twice as many atoms as 50 moles of C.
114. (E) (d) is correct.
mol
6.022 ๏ด 1023 atoms 22 electrons
91.84 g ๏ด
๏ด
๏ด
๏ฝ 2.542 ๏ด 10 25 electrons
47.867 g
mol
Ti atom
115. (M)
1 mol Fe
๏ฝ 0.0417 mol Fe
55.8452 g Fe
1 mol O
๏ฝ 0.0625 mol O
1.000 g O ๏ด
15.999 g O
Dividing the two mole values to obtain the mole ratio, we get: 0.0625/0.0417 = 1.50. That is,
1.50 moles (or atoms) of O per 1 mole of Fe, or 3 moles of O per 2 moles of Fe (Fe2O3).
Performing the above calculations for a compound with 2.618 g of Fe to 1.000 g of O yields
0.0469 mol of Fe and 0.0625 mol of O, or a mole ratio of 1.333, or a 4:3 ratio (Fe3O4).
2.327 g Fe ๏ด
116. (D) The weighted-average atomic mass of Sr is expressed as follows:
atomic mass of Sr = 87.62 amu = 83.9134(0.0056) + 85.9093x
+ 86.9089[1 โ (0.0056 + 0.8258 + x)] + 87.9056(0.8258)
Rearrange the above equation and solve for x, which is 0.095 or 9.5%, which is the relative
abundance of 86Sr. Therefore, the relative abundance of 87Sr is 0.0735 or 7.3%.
The reason for the imprecision is the low number of significant figures for 84Sr.
117. (M) This problem lends itself well to the conversion pathway:
0.15 mg Au 1 ton seawater
1 kg
1.03 g seawater 250 mL
๏ด
๏ด
๏ด
๏ด
1 ton seawater
1000 kg
1000 mg 1 mL seawater sample
1 g Au
1 mol Au
6.02 ๏ด 1023 atoms
๏ด
๏ด
๏ด
๏ฝ 1.2 ๏ด 1014 atoms of Au
1000 mg Au 196.97 g Au
1 mol Au
118. (M) In sections 2-7 and 2-8, the simplest concept is the concept of mole. Mole is defined by
the number of atoms in 12 g of 12C. Other topics emanate from this basic concept. Molar
mass (weight) is defined in terms of moles, as is mole ratios. Percent isotopic abundance is
another topic defined directly by the concept of moles.
59
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